Question

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s = 6/t2, where t is measured in seconds. Find the velocity of the particle at times t = a, t = 1, t = 2, and t = 3.

Answer 1

The velocity of the particle for different times (t = a, 1, 2, and 3 seconds) is found by differentiating the displacement equation s = 6/t^2 and applying the values of t to get the velocities: v(a) = -12/a^3, v(1) = -12 m/s, v(2) = -1.5 m/s, and v(3) ≈ -0.44 m/s.

Step-by-step explanation:

The question you've asked is about finding the velocity of a particle moving in a straight line where its displacement is given by s = 6/t2, and t is the time in seconds. To find the velocity (v) at any given time, we need to take the derivative of the displacement with respect to time. So the derivative of s with respect to t gives us v = -12/t3. Let's apply this formula for t = a, 1, 2, and 3 seconds.

• For t = a: v(a) = -12/a3 m/s
• For t = 1 second: v(1) = -12/13 m/s = -12 m/s
• For t = 2 seconds: v(2) = -12/23 m/s = -12/8 m/s = -1.5 m/s
• For t = 3 seconds: v(3) = -12/33 m/s = -12/27 m/s ≈ -0.44 m/s

Note that the negative sign indicates the direction of velocity is opposite to the direction assumed as positive in the displacement equation.

Answer 2

Velocity of the particle at time t = a

`v(a)=-(12)/(a^3)`

Velocity of the particle at time t = 1

`v(1)=-12m/s`

Velocity of the particle at time t = 2

`v(2)=-1.5m/s`

Velocity of the particle at time t = 3

`v(3)=-0.44m/s`

Step-by-step explanation:

Displacement,

`s(t)=(6)/(t^2)`

Velocity is given by

`v(t)=(ds)/(dt)=(d)/(dt)\left ( (6)/(t^2)\right )=-(12)/(t^3)`

Velocity of the particle at time t = a

`v(a)=-(12)/(a^3)`

Velocity of the particle at time t = 1

`v(1)=-(12)/(1^3)=-12m/s`

Velocity of the particle at time t = 2

`v(2)=-(12)/(2^3)=-1.5m/s`

Velocity of the particle at time t = 3

`v(3)=-(12)/(3^3)=-0.44m/s`